Suppose the nine players are A,B,C,D,E,F,G,H,I
Suppose A and B cannot be on the same team. There are two cases.
Case 1. Neither A nor B is the umpire.
A is on one team and B is on the other. Then we choose the umpire from the
7 players C,D,E,F,G,H,I. That's 7 ways to pick the umpire.
We have 6 players left. We choose 3 to play on the team with A in 6C3 = 20
ways, and the remaining 3 will play on the team with B in 3C3 = 1 way.
That's 7∙20∙1 = 140 ways for case 1.
Case 2. A or B is the umpire.
We choose the umpire 2 ways.
Then we have 8 players choose 4 on one team in 8C4 = 70 ways, and that
leaves 4C4 = 1 way to put the others on the other team.
That's 2∙70 = 140 ways for case 2.
So that's 140 for case 1 and 140 for case 2, a total of 280 ways.
Edwin
